Halve the x-coefficient, square it — and the vertex stops being hidden.
The given equation defines a parabola in the xy-plane. What is the y-coordinate of the vertex of the parabola?
y = x² + 14x + 40
Half of 14 is 7 and 7² = 49, so y = (x² + 14x + 49) − 49 + 40 = (x + 7)² − 9. The vertex is (−7, −9), and its y-coordinate is −9. C answers with the x-coordinate — the question asked for y. A stops at −49, forgetting to add the original 40 back. D reads off the constant term 40, which is the y-intercept, not the vertex.
That single instruction is the whole method: the number you need is (b2)².
b
14
halve it
7
square it
49
x² + 14x + 49 really is a perfect square, and the picture shows why: it is a square of side x + 7 with a 7 × 7 corner filled in.
The 49 is literally the missing corner — which is where the name “completing the square” comes from. Two rectangles of 7x sit along the edges, and 7x + 7x = 14x is exactly the middle term you started with.
You are rewriting an expression, not solving an equation, so the line has to stay equal to what it was.
y = x² + 14x + 40
the given equation
y = (x² + 14x + 49) − 49 + 40
+49 and −49 cancel, so nothing changed
y = (x + 7)² − 49 + 40
the bracket holds the half, 7 — not the 49
y = (x + 7)² − 9
−49 + 40 = −9
If the item hands you an expression or a function, add and subtract on the same line, as above. If it hands you an equation to solve, you may instead add 49 to both sides: x² + 14x = −40 becomes x² + 14x + 49 = 9, so (x + 7)² = 9 and x = −4 or x = −10. What you must never do is add 49 to one line and walk away — that quietly replaces the question with a different one.
Check the expansion if you doubt it: (x + 7)² − 9 = x² + 14x + 49 − 9 = x² + 14x + 40 ✓. And the sign inside the bracket copies the sign of b — b = 14 gives (x + 7)², b = −14 would give (x − 7)².
Read h with its sign flipped and k exactly as written. (x + 7)² − 9 means h = −7 and k = −9.
| In a(x − h)² + k | what it gives you | for (x + 1)² − 4 |
|---|---|---|
| h | x-coordinate of the vertex | −1 |
| k | y-coordinate of the vertex | −4 |
| x = h | the axis of symmetry | x = −1 |
| sign of a | a > 0 opens up, so k is the minimum | minimum −4 |
The dashed line is the axis of symmetry, x = −1. It runs through the vertex, and the two x-intercepts sit at equal distances from it — here −3 and 1, each two units away. Standard form x² + 2x − 3 hides all of that; vertex form (x + 1)² − 4 puts it on the page.
Halve-and-square only works on a bare x². Take a outside a bracket, complete the square inside, then multiply back out.
y = 3x² + 12x + 5
y = 3(x² + 4x) + 5
3 out of the x terms only — the 5 stays put
x² + 4x = (x + 2)² − 4
half of 4 is 2, squared is 4
y = 3[(x + 2)² − 4] + 5
put it back inside the bracket
y = 3(x + 2)² − 12 + 5
the 3 multiplies the −4 too
y = 3(x + 2)² − 7
vertex (−2, −7)
Writing 3(x + 2)² − 4 + 5 = 3(x + 2)² + 1 is the standard slip: the −4 lives inside a bracket that the 3 is multiplying, so it becomes −12. Test it at x = 0. The original gives 5. The correct form gives 3(4) − 7 = 5 ✓, and the wrong one gives 3(4) + 1 = 13 ✗. One substitution catches it every time.
The a out front never changes where the vertex sits sideways, only how steep the curve is and, if it is negative, which way it opens. With a < 0 the parabola opens downward and k becomes the maximum value instead of the minimum.
The axis of symmetry always sits at x = −b2a. Get that x, substitute it back, and you have the vertex without completing anything.
y = x² + 14x + 40, so a = 1 and b = 14
back to the opening problem
x = −14 ÷ (2 · 1) = −7
the axis of symmetry
y = (−7)² + 14(−7) + 40
substitute, calculator allowed
y = 49 − 98 + 40 = −9
same vertex, two lines
It works for step 4 too: a = 3, b = 12, so x = −12 ÷ 6 = −2 and y = 3(4) + 12(−2) + 5 = 12 − 24 + 5 = −7. Vertex (−2, −7), exactly as before.
The Digital SAT gives you a graphing calculator the whole time. Type y = x² + 14x + 40 and the vertex appears as a grey dot you can tap; the readout says (−7, −9). That is the fastest route of all when the question wants a coordinate, a minimum value or the number of x-intercepts — but it cannot answer “which equivalent form displays the vertex as constants”, because that one wants algebra.
Two tools, and the stem tells you which one it wants.
| What the stem says | Fastest route |
|---|---|
| “the y-coordinate of the vertex” | x = −b2a, then substitute |
| “the minimum value of f” / “the maximum value” | x = −b2a, then substitute |
| “for what value of x is f at its minimum” | x = −b2a, and stop there |
| “an equivalent form in which the coordinates of the vertex appear as constants” | complete the square — the answer is the form |
| “the graph has a minimum at (h, k), what is a?” | write a(x − h)² + k, then use one other point |
In the opening problem the answer −7 is on the list precisely because it is the x-coordinate, and it is the one people pick when they rush. Read the last six words of the stem again before you choose: value of f means the output, value of x means the input.
The function f is defined by f(x) = x² + bx + 20, where b is a positive constant. If the minimum value of f is 4, what is the value of b?
f(x) = (x + b2)² − b²4 + 20
halve and square, with a letter
minimum = 20 − b²4
the square is never negative, so it bottoms out at 0
20 − b²4 = 4
the condition in the stem
b²4 = 16, so b² = 64
two candidates, ±8
b = 8
“positive constant” throws out −8
“Where a is a positive constant” is never decoration. It is there because the algebra produces two answers and the stem is telling you which one to keep.
The same line of reasoning is why a square is so useful: (x + 7)² ≥ 0 for every real x, so (x + 7)² − 9 can never dip below −9. Vertex form does not just show the minimum, it proves it.
The function f is defined by the given equation. What is the maximum value of f?
f(x) = −2x² + 20x − 43
a = −2 is negative, so the parabola opens downward and the vertex is the maximum. x = −20 ÷ (2 · −2) = 5, and f(5) = −2(25) + 100 − 43 = 7. Completing the square agrees: −2(x² − 10x) − 43 = −2[(x − 5)² − 25] − 43 = −2(x − 5)² + 50 − 43 = −2(x − 5)² + 7. B answers with the x-coordinate, not the value of f. D stops at +50 and forgets the −43. A reads off the constant term, which is f(0), not the maximum.
Hint: check the sign of a before anything else — a negative a means you are looking for a maximum, and the vertex is still the place to find it. Then decide which the question wants, the input or the output.
